Difference between revisions of "2011 AMC 10A Problems/Problem 22"
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Adding all those combinations up, we get <math>600+1920+600=\boxed{3120 \ \mathbf{(C)}}</math>. | Adding all those combinations up, we get <math>600+1920+600=\boxed{3120 \ \mathbf{(C)}}</math>. | ||
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| + | == See Also == | ||
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| + | {{AMC10 box|year=2011|ab=A|num-b=21|num-a=23}} | ||
Revision as of 09:45, 8 May 2011
Problem 22
Each vertex of convex pentagon 
is to be assigned a color. There are 
colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
Solution
Let vertex 
be any vertex, then vertex 
be one of the diagonal vertices to , 
be one of the diagonal vertices to , and so on. We consider cases for this problem.
In the case that has the same color as , 
has a different color from and so 
has a different color from and . In this case, has choices, has 
choices (any color but the color of ), has 
choice, has choices, and has 
choices, resulting in a possible of 
combinations.
In the case that has a different color from and has a different color from , has choices, has choices, has choices (since and necessarily have different colors), has choices, and has choices, resulting in a possible of 
combinations.
In the case that has a different color from and has the same color as , has choices, has choices, has choices, has choice, and has choices, resulting in a possible of 
combinations.
Adding all those combinations up, we get 
.
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
